Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 10009;
int x[N], y[N], a[33], b[33];
vector<pair<int, int>> u, v;
int main() {
  int n, i, j;
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) {
    scanf("%d", x + i);
    for (j = 29; ~j; --j)
      if (x[i] >> j & 1) {
        if (!a[j]) {
          a[j] = i;
          break;
        }
        x[i] ^= x[a[j]], u.push_back({i, a[j]});
      }
  }
  for (i = 1; i <= n; ++i) {
    scanf("%d", y + i);
    for (j = 29; ~j; --j)
      if (y[i] >> j & 1) {
        if (!b[j]) {
          b[j] = i;
          break;
        }
        y[i] ^= y[b[j]], v.push_back({i, b[j]});
      }
  }
  for (i = 29; ~i; --i)
    if (b[i]) {
      for (y[b[i]] ^= x[a[i]], j = i - 1; ~j; --j)
        if (y[b[i]] >> j & 1) y[b[i]] ^= x[a[j]], u.push_back({a[i], a[j]});
      if (y[b[i]]) puts("-1"), exit(0);
    } else if (a[i])
      u.push_back({a[i], a[i]}), a[i] = 0;
  for (i = 29; ~i; --i)
    if (b[i] && a[i] != b[i]) {
      u.push_back({a[i], b[i]}), u.push_back({b[i], a[i]}),
          u.push_back({a[i], b[i]});
      for (j = 29; ~j; --j)
        if (a[j] == b[i]) a[j] = a[i];
    }
  cout << u.size() + v.size() << '\n';
  for (auto i : u) printf("%d %d\n", i.first, i.second);
  for (auto i = v.rbegin(); i != v.rend(); ++i)
    printf("%d %d\n", i->first, i->second);
  return 0;
}